The point P = (x, -) lies on the unit circle shown below. What is the value of r in simplest form? %3D (1, o) P (x, y)
The point P = (x, -) lies on the unit circle shown below. What is the value of r in simplest form? %3D (1, o) P (x, y)
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![### Determining the Value of \( x \)
**Question:**
The point \( P = \left( x, -\frac{2}{7} \right) \) lies on the unit circle shown below. What is the value of \( x \) in simplest form?
**Diagram Explanation:**
The diagram depicts a unit circle centered at the origin \((0, 0)\) on a coordinate plane. The unit circle has a radius of 1. The point \( P(x, y) \) is marked on the circumference of the circle in the fourth quadrant with coordinates \( P \left( x, -\frac{2}{7} \right) \). The x-axis and y-axis intersect at the center of the circle.
**Note:** The figure is not drawn to scale.
**Solution:**
To find the value of \( x \) when a point lies on a unit circle, we use the Pythagorean identity for a point \((x, y)\) on a unit circle which states that \( x^2 + y^2 = 1 \).
Given:
\[ y = -\frac{2}{7} \]
Substitute \( y \) into the Pythagorean identity:
\[ x^2 + \left( -\frac{2}{7} \right)^2 = 1 \]
\[ x^2 + \frac{4}{49} = 1 \]
Subtract \( \frac{4}{49} \) from both sides:
\[ x^2 = 1 - \frac{4}{49} \]
\[ x^2 = \frac{49}{49} - \frac{4}{49} \]
\[ x^2 = \frac{45}{49} \]
Take the square root of both sides:
\[ x = \pm \sqrt{ \frac{45}{49} } \]
\[ x = \pm \frac{ \sqrt{45} }{7} \]
Simplify \( \sqrt{45} \):
\[ \sqrt{45} = \sqrt{9 \times 5} = 3 \sqrt{5} \]
Therefore:
\[ x = \pm \frac{3 \sqrt{5}}{7} \]
So, the value of \( x \) in simplest form is:
\[ x = \pm \frac{3 \sqrt{5}}{7} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49e7cf44-c454-4b76-b99a-274946cc3378%2F9a137d27-5a95-477b-acec-9403aeacb04c%2F9n1fmbk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining the Value of \( x \)
**Question:**
The point \( P = \left( x, -\frac{2}{7} \right) \) lies on the unit circle shown below. What is the value of \( x \) in simplest form?
**Diagram Explanation:**
The diagram depicts a unit circle centered at the origin \((0, 0)\) on a coordinate plane. The unit circle has a radius of 1. The point \( P(x, y) \) is marked on the circumference of the circle in the fourth quadrant with coordinates \( P \left( x, -\frac{2}{7} \right) \). The x-axis and y-axis intersect at the center of the circle.
**Note:** The figure is not drawn to scale.
**Solution:**
To find the value of \( x \) when a point lies on a unit circle, we use the Pythagorean identity for a point \((x, y)\) on a unit circle which states that \( x^2 + y^2 = 1 \).
Given:
\[ y = -\frac{2}{7} \]
Substitute \( y \) into the Pythagorean identity:
\[ x^2 + \left( -\frac{2}{7} \right)^2 = 1 \]
\[ x^2 + \frac{4}{49} = 1 \]
Subtract \( \frac{4}{49} \) from both sides:
\[ x^2 = 1 - \frac{4}{49} \]
\[ x^2 = \frac{49}{49} - \frac{4}{49} \]
\[ x^2 = \frac{45}{49} \]
Take the square root of both sides:
\[ x = \pm \sqrt{ \frac{45}{49} } \]
\[ x = \pm \frac{ \sqrt{45} }{7} \]
Simplify \( \sqrt{45} \):
\[ \sqrt{45} = \sqrt{9 \times 5} = 3 \sqrt{5} \]
Therefore:
\[ x = \pm \frac{3 \sqrt{5}}{7} \]
So, the value of \( x \) in simplest form is:
\[ x = \pm \frac{3 \sqrt{5}}{7} \]
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